Q. 11. Calculate the mass of NaBr needed to prepare a 50.00-mL aqueous solution that would yield the same conductivity as the NaCl (aq) solution. Show your calculation to your instructor before making the solution and measuring its conductivity.

The mass of NaBr needed is 0.22969 g.

Explanation:

1 mole of NaBr contains 22.4 dm^3 of NaBr

Therefore, 0.05 dm^3 (50/1000 = 0.05 dm^3) of NaBr = 0.05/22.4 = 0.00223 mol of NaBr

MW of NaBr = 23 + 80 = 103 g/mol

Mass of NaBr = number of moles * MW = 0.00223 * 103 = 0.22969 g